REMARK ON THE OPTIMAL REGULARITY FOR EQUATIONS OF WAVE MAPS TYPE

REMARK ON THE OPTIMAL REGULARITY FOR EQUATIONS OF WAVE MAPS TYPE Sergiu Klainerman and Sigmund Selberg 1 Introduction Department of Mathematics Princeton University Princeton, NJ 08544, USA The goal of this paper is to review the estimates proved in [3] and extend them to all dimensions, in particular to the harder case of space dimension. As in [3], the main application we have in view is to equations of Wave Maps type, namely systems of equations of the form (1) φ I + Γ I JK (φ)q 0(φ J, φ K ) = 0. Here, = t + denotes the standard D Alembertian in Rn+1, and Q 0 is the null form Q 0 (φ, ψ) = α φ α ψ = t φ t ψ + n i φ i ψ. The main estimates of the paper are described in Theorems and 3. It is easy to see, using the arguments in [3] and [5], that these estimates imply the following result: i=1

Theorem 1. Assume that the functions Γ(φ) are real analytic. Then the initial value problem for the equations (1), in R n+1, subject to the initial conditions () φ(0, x) = f 0 (x), t φ(0, x) = g 0 (x), is well posed for f 0 H s (R n ), g 0 H s 1 (R n ) and any s > n. In particular, in two space dimensions, our result comes very close to proving well posedness in the energy norm H 1. In [1] the above Theorem was proved in the special case of dimension n = 3. The proof in higher dimensions does not require any new idea. The case of dimension n = was studied by Y. Zhou [6], who showed well posedness for H s, s 1 + 1 8. Here we rely on a slight modification of a new version of Strichartz type estimates proved in [4]. As in [3], the proof relies on estimates for the null quadratic form Q 0 in the spaces H s,δ. The estimates diverge logarithmically for the critical exponents s = n, δ = 1. In this paper we keep track of the precise divergences by using the modified spaces H [s,δ] which will be discussed below. Estimates involving the homogeneous 1 version of these norms, for the optimal exponents δ = ± 1 in R3+1, have first appeared in []. For example section 6 in [] proves, essentially, the estimates included here in Theorem 4 for the particular case of space dimension 3, and applies them to derive the following sharper version of estimate (4) of Theorem in the case when φ = ψ = 0 with initial data bounded in H 3 (R 3 ): ( τ + ξ ) 1 τ ξ 1 Q0 (φ, ψ) <. L (R 3+1 ) The inhomogeneous version of the norms, which have appeared in [3] and [5], was inspired from Bourgain s work [1]. The inhomogeneous norms appear 1 i.e., with w ± (τ, ξ) replaced by τ ± ξ.

naturally in connection with Bourgain s time cut-off idea, which allows one, essentially, to replace the symbol τ ξ of by w + w (τ, ξ) and thus circumvent the difficulties which appear when trying to treat the sharp local well posedness in H n. That problem remains in fact open and probably requires to go back, in some way, to homogeneous estimates. Notation We define the space H [s,δ] (R 1+n ) to be the completion of the Schwartz space S(R 1+n ) w.r.t. the norm φ H [s,δ] = w+ s L(w +)(τ, ξ)w δ L(w )(τ, ξ) φ(τ, ξ), L τ,ξ where φ is the space-time Fourier transform of φ, the weights w ± are given by w ± (τ, ξ) = + τ ± ξ and L : (0, ) (0, ) is a function with the properties L is increasing, L(x) CL(x) for all x, dx <. xl (x) Note that the first two properties imply that for any a > 1, (3) L(ax) C a L(x) for all x. Examples of such a function L are the following: L(x) = x ε, for any ε > 0,

L(x) = log x, L(x) = (log x) 1 log log x. Introducing the notation w [s] + = w s +L(w + ) w [δ] = w δ L(w ), we have φ H [s,δ] = w [s] + w [δ] φ L. The symbol will mean up to a multiplicative constant, which may depend on the space dimension n and other fixed parameters, but not on any variable quantities. Similarly, means = up to a constant. 3 The Main Estimates Theorem. Assume that s n and n. Then (4) Q 0 (φ, ψ) H [s 1, 1 ] φ H [s, 1 ] ψ H [s, 1 ]. Also, if P is a polynomial in k variables, P(x) = α N a αx α, then (5) P(φ 1,..., φ k )Q 0 (φ, ψ) H [s 1, 1 ] Cp(C φ 1 H [s, 1 ],..., C φ k H [s, 1 ] ) φ H [s, 1 ] ψ H [s, 1 ], where p(x) = α N a α x α and C is a constant independent of P. Theorem 3. Assume that s n and n. Then (6) φψ H [s, 1 ] φ H [s, 1 ] ψ H [s, 1 ] and (7) φψ H [s 1, 1 ] φ H [s 1, 1 ] ψ H [s, 1 ].

We will show that Theorem follows from Theorem 3, which in turn can be reduced to the following two bilinear estimates for solutions of the homogeneous wave equation. Theorem 4. Let φ and ψ be solutions of the homogeneous wave equation on R 1+n, n,with data φ(0, ) = f, t φ(0, ) = 0 and ψ(0, ) = g, t ψ(0, ) = 0. Then the following estimates hold: 1 τ ξ L (8) φψ(τ, ξ) f L (R n ) g Ḣ n (R n ) τ,ξ 1 τ ξ (9) ( ) n φψ(τ, ξ) f L τ + ξ (R n ) g Ḣ 1 (R n ). L τ,ξ Here, f Ḣs = ξ s ˆf(ξ) L ξ is the homogeneous Sobolev norm of exponent s. To see that Theorem follows from Theorem 3, first notice that (10) Q 0 (φ, ψ)(τ, ξ) q 0 (τ λ, ξ η; λ, η) φ(τ λ, ξ η) ψ(λ, η) dλ dη, where q 0 is the symbol of the null form Q 0, i.e., q 0 (τ, ξ; λ, η) = τλ ξ η. [ From the identity τλ ξ η = 1 (τ + λ) ξ + η τ + ξ λ + η ], it follows that q 0 (τ, ξ; λ, η) w + w (τ + λ, ξ + η) + w + w (τ, ξ) + w + w. Combined with (10), and since we may assume w.l.o.g. that φ and ψ are both non-negative, this shows that Q 0 (φ, ψ) H [s 1, 1 ] φψ H [s, 1 + (Λ ] + Λ φ)ψ H [s 1, 1 + φ(λ ] + Λ ψ) H [s 1, 1, ] where the operators Λ ± are defined by Λ ± φ = w ± φ. Thus, (4) follows from Theorem 3, while (5) follows by an induction argument using (4) and (7).

4 Proof of Theorem 4 We will need two lemmas. Lemma 1. If η 1 and η are two points on the ellipsoid η + ξ η = τ, where τ > ξ,ξ R n, then η 1 η η1 η ( τ ξ )1. Proof. If η 1 η ( τ ξ ) 1, this is clearly true, so we assume η 1 η ( τ ξ ) 1. Then η 1 η η1 η η 1 η η1 η = η 1 η + η1 η ( ) η 1 η η 1 η ( ) η 1 η η 1 η η 1 η (. τ ξ )1 But in [4] it was proved that η 1 η η 1 η τ ξ, and we are done. Lemma. Assume n and a, b R. The estimate ( τ ξ ) ( ) n 3 δ τ η ξ η dη τ a b η a ξ η b holds for all (τ, ξ) with τ > ξ if and only if a, b < n+1. Proof. We denote the above integral by I. Introducing polar coordinates η = ρω, we calculate I ( τ ξ ) n 3 Sn 1 ρ τ ξ ρn 3 ρ 1 a (τ ρ) 1 b dσ S n 1(ω), where ρ = ρ(τ, ξ, ω) = τ ξ τ. Introducing the notation λ =, y = ξ (τ ξ ω) ξ and x = ρ ξ = λ 1 (λ y), we have dx dy = λ 1 (λ y), and hence I ( τ ξ ) n 3 ξ a b+n 3 S n 1 dx dy xn 3 x 1 a (λ x) 1 b dσ S n 1(ω) ( τ ξ ) n 3 ξ a b+n 3 1 1 ξ ω dx dy xn 3 x 1 a (λ x) 1 b (1 y ) n 3 dy,

where the last step follows from the formula f( ξ S ξ ω) dσ S n 1(ω) = dσ S n (ω ) n 1 S n 1 Noting that x x y = (λ 1) ( 1 4 (x λ )), we finally get λ+1 I ξ a b λ 1 ξ a b 1 and it is clear that 1 1 1 1 f(y)(1 y ) n 3 dy. ( ( x 1 a (λ x) 1 b 1 4 x λ ) ) n 3 dx (λ + x) 1 a (1 + x) n 3 (λ x) 1 b (1 x) n 3 dx, (λ + x) 1 a (1 + x) n 3 (λ x) 1 b (1 x) n 3 dx λ a b for all λ > 1 if and only if a, b < n+1. To prove (8), it is enough to prove that it holds with φψ replaced by φ + ψ ± = φ + ψ ±, where φ + (τ, ξ) δ ( τ ξ ) ˆf(ξ) ψ ± (λ, η) δ ( λ η ) ĝ(η). Note that φ + ψ ± (τ, ξ) = φ + ψ ± (τ, ξ) δ ( τ η ξ η ) F(ξ η) G(η) dη, η n where F(ξ) = ˆf(ξ) and G(η) = η n ĝ(η). We estimate φ + ψ + first. Applying the Cauchy-Schwarz inequality w.r.t. the measure δ ( τ η ξ η ) dη, we get τ ξ φ + ψ + (τ, ξ) ( ) τ ξ δ τ η ξ η η n dη δ ( τ η ξ η ) F (ξ η) G (η) dη,

so it suffices to show that (11) ( τ ξ ) δ ( τ η ξ η ) η n dη 1 when τ > ξ. Introducing polar coordinates, (11) can be written τ ξ n 1ρ(τ ρ) (1) τ ξ dσ ρ n S n 1(ω) S n 1 ρ where ρ = τ ξ. Noting that 0 < ρ < τ, we see that (1) is indeed (τ ξ ω) bounded independently of (τ, ξ). To estimate φ + ψ, we write φ + ψ = A + B, where A(τ, ξ) = δ ( τ + η ξ η ) F(ξ η) G(η) dη η < ξ η n B(τ, ξ) = δ ( τ + η ξ η ) F(ξ η) G(η) dη. η ξ η n By applying the Cauchy-Schwarz inequality as above, the estimate for A is reduced to showing that ( ξ τ ) η < ξ δ ( τ + η ξ η ) η n dη 1 when τ < ξ. This integral can be written ξ τ ξ ρ n 1ρ(τ + ρ) dσ τ ρ n S n 1(ω), S n 1 {0<ρ< ξ } where now ρ = ξ τ, and this integral is obviously bounded independently (τ+ξ ω) of (τ, ξ). To estimate B, note that if η ξ, then ξ η η. Thus, we have B(τ, ξ) δ ( τ + η ξ η ) F (ξ η) G (η) ξ η n n dη. 4 η 4 See [3].

Denoting this last integral by I, what we have to show is that (13) ( ξ τ ) I dτ dξ F L G L. To prove this, we use an argument introduced in [4]. First, we write I = i=1 δ ( τ + η i ξ η i ) F (ξ η i ) G (η i ) ξ η i n 4 ηi n 4 Multiplying by ξ τ and integrating w.r.t. dτ dξ, we get ( ξ τ ) I dτ dξ dη i. ( ξ ξ η1 η 1 ) δ ( ξ η 1 η 1 + η ξ η ) Now we change variables This gives F (ξ η 1) G (η 1 ) F (ξ η ) G (η ) ξ η 1 n 4 η1 n 4 ξ η n 4 η n 4 (ξ, η 1, η ) (ξ, η 1, η ) ξ = ξ η 1 η, η 1 = η, η = η 1. dη 1 dη dξ. ξ = ξ η 1 η ξ η 1 = ξ η 1 ξ η = ξ η, so the last integral is transformed to, when we drop the primes on the new variables, ( ξ η 1 η ξ η1 η ) δ ( ξ η 1 + η 1 η ξ η ) F (ξ η 1) G ( η 1 ) F (ξ η ) G ( η ) ξ η 1 n 4 η1 n 4 ξ η n 4 η n 4 dη 1 dη dξ.

Applying Cauchy-Schwarz w.r.t. the measure δ ( ξ η 1 + η 1 η ξ η ) dη 1 dη dξ, we bound this by ( ( ξ η 1 η ξ η1 η ) δ ( ξ η 1 + η 1 η ξ η ) F (ξ η 1 ) G ( η 1 ) dη ξ η n η n 1 dη dξ ( ( ξ η 1 η ξ η1 η ) δ ( ξ η 1 + η 1 η ξ η ) )1 F (ξ η ) G ( η ) ξ η 1 n η1 n dη 1 dη dξ )1. Thus, the estimate (13) has been reduced to the following lemma, which is an immediate consequence of Lemma 1 and Lemma. Lemma 3. If n, the integral ( η η 0 η η0 ) δ ( τ η ξ η ) I(τ, ξ, η 0 ) = ξ η n n dη η is bounded uniformly in (τ, ξ, η 0 ), where τ > ξ and η 0 + ξ η 0 = τ. Finally, we consider (9). If n =, the estimates (8) and (9) coincide, so we assume n 3. Proceeding as in the proof of (8), the proof of (9) can be reduced to an estimate ( ξ τ ) ( τ + ξ ) n I dτ dξ F L G L, where now I = δ ( τ + η ξ η ) F(ξ η) G(η) dη. ξ η 1 η 1

In other words, we have to prove the estimate ( ) 1 ξ τ ( ) n φ + ψ f Ḣ 1 τ + ξ L g (R n ) Ḣ 1. (R n ) τ,ξ In fact, something stronger is true. In [4] it was proved that 1 ( ) n 3 φ + ψ f Ḣ 1 τ + ξ L g (R n ) Ḣ 1 (R n ) τ,ξ when n 3. 5 Proof of Theorem 3 We first prove (6). Writing F(τ, ξ) = w [s] + w [ 1 ] (τ, ξ) φ(τ, ξ) G(λ, η) = w [s] + w [ 1 ] ψ(λ, η), we have [s] w + w [ 1 ] (τ, ξ)f(τ λ, ξ η)g(λ, η) φψ H [s, 1 = dλ dη ] w [s] + w [ 1 ] (τ λ, ξ η)w [s] + w [ 1 ] so by the self-duality of L, it will be enough to prove that L τ,ξ, (14) w [s] + w [ 1 ] (τ, ξ) F (τ λ, ξ η) G (λ, η) H (τ, ξ) w [s] + w [ 1 ] (τ λ, ξ η)w [s] + w [ 1 ] dλ dη dτ dξ F L (R 1+n ) G L (R 1+n ) H L (R 1+n ) for all F, G, H L (R 1+n ). From now on, we drop the absolute values on F, G and H, and simply assume that F, G, H 0. In order to estimate the

integral (14), we split the domain of integration into several pieces. First, note that by symmetry, we may assume w + (λ, η) w + (τ λ, ξ η). Combining this with the inequality (15) w + (τ, ξ) w + (λ, η) + w + (τ λ, ξ η), and making use of the property (3) of the function L, we get so we are left with the integral w [s] + (τ, ξ) w [s] + (τ λ, ξ η), (16) I = w [ 1 ] (τ, ξ)f(τ λ, ξ η)g(λ, η)h(τ, ξ) w [ 1 ] (τ λ, ξ η)w [s] + w [ 1 ] dλ dη dτ dξ. To deal with the factor w [ 1 ] (τ, ξ) in the numerator, we need the following lemma, the proof of which can be found in [5]. Lemma 4. We have (17) w (τ, ξ) w (τ λ, ξ η) + w + d ± (ξ η, η), where d ± (ξ η, η) = ξ η ± η ξ, and we choose + or according to whether τ λ and λ have equal or opposite signs. Using Lemma 4, we split the domain of integration of I into three parts, according to which of the three terms on the right hand side of (17) is largest.

Using once more the property (3), we thus get I I 1 + I + I 3, where (18) (19) (0) F(τ λ, ξ η)g(λ, η)h(τ, ξ) I 1 = dλ dη dτ dξ w [s] + w [ 1 ] F(τ λ, ξ η)g(λ, η)h(τ, ξ) I = dλ dη dτ dξ w [ 1 ] (τ λ, ξ η)w [s] + (λ, η) 1 d I 3 = ±L(d ± )(ξ η, η)f(τ λ, ξ η)g(λ, η)h(τ, ξ) dλ dη dτ dξ. w [ 1 ] (τ λ, ξ η)w [s] + w [ 1 ] Now we perform the change of variables (1a) (1b) (τ, λ) (u, v) u = τ λ ξ η, v = λ η. Without loss of generality, we assume that τ λ 0. Thus, u = τ λ ξ η, v = ±λ η, and we have I 1 I where ( ) f u (ξ η)g v (η)h u,v ξ η ± η,ξ du dv dη dξ ( + v ) 1 L( + v )( + η ) s L( + η ) ( ) f u (ξ η)g v (η)h u,v ξ η ± η,ξ du dv dη dξ, ( + u ) 1 L( + u )( + η ) s L( + η ) f u (ξ) = F ( u + ξ,ξ ) g v (η) = G ( ±(v + η ), η ) H u,v (τ, ξ) = H(u ± v + τ, ξ). Therefore, using the following lemma first w.r.t. the (u, v) variables and then w.r.t. (ξ, η), we get the estimate I i F L G L H L for i = 1,.

Lemma 5. Assume a n, n 1. Then f(x)g(y)h(x + y) R n R ( + x ) n a L( + x ) dxdy f L (R n ) g L (R n ) h L (R n ). Proof. By the Cauchy-Schwarz inequality, the above integral is dominated by ( )1 dx ( g(y) ( + x ) a L ( + x ) f (x) h (x + y) dx )1 dy. By the assumptions on L, the integral in front is bounded, so applying Cauchy-Schwarz once more finishes the proof. We now turn our attention to the difficult part, I 3. We obtained I 3 after using Lemma 4 and then restricting I to the region where w (τ λ, ξ η), w d ± (ξ η, η). Note that w (τ λ, ξ η) and w measure how far removed the points (τ λ, ξ η) and (λ, η) are from the light cone through the origin. Thus, we may think of I 3 as the part of I that results from concentrating F and G on the light cone. To estimate I 3, note that by the triangle inequality, d ± (ξ η, η) η. Therefore, by the properties of L, we have L(d ± )(ξ η, η) L(w + )(λ, η), whence I 3 1 d ±(ξ η, η)f(τ λ, ξ η)g(λ, η)h(τ, ξ) w [ 1 ] (τ λ, ξ η)w [ 1 ] ( + η ) n Performing the change of variables (1), we get I 3 dλ dτ dη dξ. J u,v + + J u,v du dv, ( + u ) 1 L( + u )( + v ) L( 1 + v )

where J u,v ± = d 1 ( ) 1 ± (ξ η, η)f u (ξ η)g v (η)h u,v ξ η ± η,ξ dη dξ. η n Note that if we can show that () J ± u,v f u L (R n ) g v L (R n ) H L (R 1+n ), then the estimate I 3 F L G L H L follows after an application of the Cauchy-Schwarz inequality w.r.t. du and dv. To prove (), we rewrite the integral J u,v ± as follows: 1 τ ξ f u (ξ η)g v (η)δ ( τ η ξ η ) H u,v (τ, ξ) dη dτ dξ. η n This is dominated by 1 τ ξ fu (ξ η)g v (η)δ ( τ η ξ η ) dη H η n u,v L (R 1+n ). L τ,ξ Noting that (i) H u,v L = H L, and (ii) by the proof of (8), 1 τ ξ fu (ξ η)g v (η)δ ( τ η ξ η ) dη f η n u L g v L, L τ,ξ the proof of (6) is complete. To prove (7), it suffices to prove that (3) w [s 1] + w [ 1 ] (τ, ξ)f(τ λ, ξ η)g(λ, η)h(τ, ξ) w [s 1] + w [ 1 ] (τ λ, ξ η)w [s] + w [ 1 ] dλ dη dτ dξ F L G L H L.

First, we use (15) to dominate the integral in (3) by the sum of the two integrals (4) (5) w [ 1 ] (τ, ξ)f(τ λ, ξ η)g(λ, η)h(τ, ξ) w [ 1 ] (τ λ, ξ η)w [s] [ w 1 ] (τ, ξ)f(τ λ, ξ η)g(λ, η)h(τ, ξ) + w [ 1 ] w [s 1] + w [ 1 ] (τ λ, ξ η)w + w [ 1 ] dλ dη dτ dξ dλ dη dτ dξ. Next, using the fact that (since L is increasing) w 1 (τ λ, ξ η)l(w )(τ, ξ) w 1 (τ, ξ)l(w )(τ λ, ξ η) w 1 L(w )(τ,ξ) w 1 L(w )(τ λ,ξ η) w 1 L(w )(τ λ,ξ η) w 1 L(w )(τ,ξ) if w (τ λ, ξ η) w (τ, ξ) if w (τ, ξ) w (τ λ, ξ η), we see that (4) reduces to the integral (16), which was estimated above, while (5) is dominated by the sum of the integrals (6) (7) w [ 1 ] (τ, ξ)f(τ λ, ξ η)g(λ, η)h(τ, ξ) dλ dη dτ dξ w [s 1] + w [ 1 ] (τ λ, ξ η)w + w [ 1 ] [ w 1 ] (τ λ, ξ η)f(τ λ, ξ η)g(λ, η)h(τ, ξ) w [s 1] + (τ λ, ξ η)w [ 1 ] (τ, ξ)w + w [ 1 ] dλ dη dτ dξ. Integral (6) reduces to integrals of the type (16) when we consider separately the regions where w + (λ, η) w + (τ λ, ξ η) and w + (τ λ, ξ η) < w + (τ, ξ), so we are left with the integral (7), which after a change of variables takes the form (8) w [ 1 ] (τ, ξ)f(τ, ξ)g(λ, η)h(τ λ, ξ η) w [s 1] + (τ, ξ)w [ 1 ] (τ λ, ξ η)w + w [ 1 ] dλ dη dτ dξ.

(Here we have replaced G(λ, η) by G( λ, η).) We now consider the two regions given by (9) (30) w (τ λ, ξ η) + w 1 d ±(ξ η, η) w (τ λ, ξ η) + w < 1 d ±(ξ η, η) Using Lemma 4, we find that (8) restricted to the region (9) is dominated by the sum of the two integrals F(τ, ξ)g(λ, η)h(τ λ, ξ η) w [s 1] + (τ, ξ)w + w [ 1 ] F(τ, ξ)g(λ, η)h(τ λ, ξ η) w [s 1] + (τ, ξ)w [ 1 ] (τ λ, ξ η)w + (λ, η) dλ dη dτ dξ dλ dη dτ dξ. Both these integrals reduce to integrals of the type (18) or (19) when we consider separately the regions w + (τ, ξ) w + (λ, η) and w + (λ, η) < w + (τ, ξ). Now assume that we are in the region (30). Then by Lemma 4, (31) w [ 1 ] (τ, ξ) d 1 ± L(d ± )(ξ η, η). In fact, we also have (3) w [s 1] + (τ, ξ) D n ± L(D ± )(ξ η, η), where D ± (ξ η, η) = ξ η ± η + ξ. This follows from the estimate w + (τ, ξ) τ + ξ 1 D ±(ξ η, η), which clearly holds in the case, and which in the + case is a consequence of the following lemma and the requirement (30). Lemma 6. If τ λ and λ have the same sign, then τ ( ξ η + η ) w (τ λ, ξ η) + w.

The proof is obvious. Using (31),(3) and the fact that d ± D ±, and hence L(d ± ) L(D ± ), we have then to show that d 1 ± (ξ η, η)f(τ, ξ)g(λ, η)h(τ λ, ξ η) D n ± (ξ η, η)w [ 1 ] (τ λ, ξ η)( + η )w [ 1 ] dλ dη dτ dξ F L G L H L. Performing the change of variables (1), we find that the estimate can be reduced to proving that (33) J ± u,v F L g v L h u L, where g v (η) = G ( ±(v + η ), η ), h u (ξ) = H ( u + ξ,ξ ) and J ± u,v = d 1 ± (ξ η, η) F ( u ± v + ξ η ± η,ξ)g ( ±(v + η ), η ) D n ± (ξ η, η) η H ( u + ξ η,ξ η ) dη dξ. Setting F u,v (τ, ξ) = F(u ± v + τ, ξ), we rewrite J ± u,v as follows: 1 τ ξ δ ( τ η ξ η ) F u,v (τ, ξ)g v (η)h u (ξ η) ( ) n dη dτ dξ, τ + ξ η Applying the Cauchy-Schwarz inequality w.r.t. the measure dτ dξ, we see that (33) follows from the estimate 1 τ ξ ( ) n τ + ξ gv (η)h u (ξ η)δ ( τ η ξ η ) dη η L τ,ξ g v L h u L, which holds by the proof of (9). This concludes the proof of Theorem 3.

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